如何在php中使用两条select语句

Question

我想在我的php文件中有两条select语句，但我遇到了一些困难。

基本上，我试图从两个查询中获取细节并在我的android应用程序中使用它们。

这是JSON格式我想：

[ 
    userstype:[ 
    { 
     "usertypename" : name 

    } 
    ] 
    userdetails:[ 
    { 
     "forename" : forename 
     "surname" : surname 
     "age" : age 
    }] 
] 

这是我php文件：

<?php 
require "init.php"; 
$stmt = "SELECT userstypename FROM tbluserstype"; 
$result = mysqli_query($conn, $stmt); 
$outcome = array(); 
if(mysqli_num_rows($result)){ 
    while($row = mysqli_fetch_assoc($result)){ 
     $outcome[] = array 
     (
      "userstypename" => $row["userstypename"] 
     ); 
    } 
    echo json_encode($outcome); 
} 
$stmt2 = "SELECT forename, surname, age FROM users"; 
$result = mysqli_query($conn, $stmt2); 
$outcome = array(); 
if(mysqli_num_rows($result)){ 
    while($row = mysqli_fetch_assoc($result)){ 
     $outcome[] = array 
     (
      "forename" => $row["forename"], 
      "surname" => $row["surname"], 
      "age" => $row["age"] 
     ); 
    } 
    echo json_encode($outcome); 
} 
else{ 
    echo json_encode("Failed"); 
} 

?> 

我想用userstype和userdetails作为标签

Answer 1

<?php 
    require "init.php"; 
    $stmt = "SELECT userstypename FROM tbluserstype"; 
    $result = mysqli_query($conn, $stmt); 
    $outcome = array(); 
    if(mysqli_num_rows($result)){ 
    while($row = mysqli_fetch_assoc($result)){ 
    // following line modified slightly 
     $outcome['userstype'] = array(
      "userstypename" => $row["userstypename"] 
     ); 
    } 
    // take this line outecho json_encode($outcome); 
    } 
    $stmt2 = "SELECT forename, surname, age FROM users"; 
    $result = mysqli_query($conn, $stmt2); 
    //$outcome = array(); <- take this out 
    if(mysqli_num_rows($result)){ 
     while($row = mysqli_fetch_assoc($result)){ 
     // following line modified slightly 
      $outcome['userdetails'] = array(
      "forename" => $row["forename"], 
      "surname" => $row["surname"], 
      "age" => $row["age"] 
     ); 
     } 
    //take this out too, move it outside of if loop echo  json_encode($outcome); 
    }else{ 
    echo json_encode("Failed"); 
    } 
    echo json_encode($outcome); 

？ >

Answer 2

我希望这个解决方案能帮助你生成你的问题中定义的json格式。

<?php 
require "init.php"; 
$stmt = "SELECT userstypename FROM tbluserstype"; 
$result = mysqli_query($conn, $stmt); 
$outcome = array(); 
if(mysqli_num_rows($result)){ 
    while($row = mysqli_fetch_assoc($result)){ 
     // following line modified slightly 
     $outcome['userstype'] = array(
      "userstypename" => $row["userstypename"] 
     ); 
    } 
    // take this line outecho json_encode($outcome); 
} 
$stmt2 = "SELECT forename, surname, age FROM users"; 
$result = mysqli_query($conn, $stmt2); 
//$outcome = array(); <- take this out 
if(mysqli_num_rows($result)){ 
    while($row = mysqli_fetch_assoc($result)){ 
     // following line modified slightly 
     $outcome['userdetails'] = array(
      "forename" => $row["forename"], 
      "surname" => $row["surname"], 
      "age" => $row["age"] 
     ); 
    } 
    //take this out too, move it outside of if loop echo    json_encode($outcome); 
    }else{ 
     echo json_encode("Failed"); 
    } 
    // to represent your outcome with json array, write this line. 
    $mergedArray[] = $outcome; 
    echo json_encode($mergedArray); 
?>